求Lim(x->0)(sinx-e^x+1)/[1-(1-x^2)^1/2] 的极限

J=(sinx - e^x +1)/[1-(1-x^2)^0。5]用洛必达法则：lim(x->0)J=lim(x->0) (cosx - e^x)/[x/(1-x^2)^0。5] =lim(x->0) (cosx - e^x)(1-x^2)^0。5/x //:还得用一次洛必达法则=lim(x->0) [(-sinx - e^x)(1-x^2)^0。5+(cosx - e^x)*0。5*(-2x)/(1-x^2)^0。5]=-1+0=-1